Question

Current Attempt in Progress A player of a video game is confronted with a series of 4 opponents and a(n) 75% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends). Round your answers to 4 decimal places. (a) What is the probability that a player defeats all 4 opponents in a game? (b) What is the probability that a player defeats at least 2 opponents in a game? (c) If the game is played 3 times, what is the probability that the player defeats all 4 opponents at least once?

Answer 1

Answer:

Given that:

A player of a video game is confronted with a series of 4 opponents and a(n) 75% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends).

Here we have  

n = 4, p = 0.75

a) What is the probability that a player defeats all 4 opponents in a game?

12 n-2 P(= 4) Cup" (1 p)"

P(x = 4) =+ C4* (0.75)*(1 – 0.75)4-4

P(x = 4) = C4 * 0.75 0.754*0.25

Pr = 4) = 1* 0.3164 * 1

P1 = 4) = 0.3164

b) What is the probability that a player defeats at least 2 opponents in a game?

PC > 2) = 1- Px < 2)

P(> 2) = 1- [P(x = 0) + P(x = 1)

Pic > 2) = 1-[*60* (0.75)º(1 – 0.75)4-0 +4 C1 * (0.75)'(1 – 0.75)4-1

PCI > 2) = 1- [1*1 * 0.0039 +4*0.75 * 0.0156

PCI > 2) = 1- (0.0039 + 0.0468]

PC > 2) = 1-0.0507

PC > 2) = 0.9493

c) If the game is played 3 times, what is the probability that the player defeats all 4 opponents at least once?

PC > 1) = 1-PC < 1)

PC > 1) = 1- Px = 0)

P(1 > 1) = 1 _3.co * (0.3164)'(1 – 0.3164)3-

PX > 1) = 1 _3 Со * 0.3164° * * 0.6836

P(x21) = 1 – 1 * 1 * 0.3194

P(x21) = 1 – 0.3194

PC > 1) = 0.6806